a plane wall of thickness 10 cm and thermal conductivity of 25w/m degree centigrade has a volumetric heat generation of.3*10 to the power 16w/mque .the wall is insulated on one side & other side is exposed to fluid at90 degree temperature . determine maximum temperature in wall if convective heat transfer coefficient between wall & fluid is 500w/msqarek
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To determine the maximum temperature in the wall, we need to consider the heat conduction through the wall and the heat transfer from the fluid to the wall.
Let's denote:
- Thickness of the wall: L = 10 cm = 0.1 m
- Thermal conductivity of the wall: k = 25 W/m°C
- Volumetric heat generation: Q = 3 x 10^16 W/m^3
- Temperature of fluid: T_f = 90 °C
- Convective heat transfer coefficient: h = 500 W/m^2°C
First, let's calculate the heat conduction through the wall using Fourier's law of heat conduction. The rate of heat transfer through the wall can be expressed as:
Q_conduction = (k * A * ΔT) / L
Where:
- Q_conduction is the heat conduction rate through the wall.
- A is the cross-sectional area of the wall.
- ΔT is the temperature difference across the wall (T_wall - T_f).
Since the wall is insulated on one side, the cross-sectional area of the wall is the same as the exposed side. Thus:
A = L * 1 m = 0.1 m²
We need to solve for ΔT, which is the temperature difference between the wall and the fluid. We can rewrite the equation as:
ΔT = (Q_conduction * L) / (k * A)
Now, let's calculate Q_conduction:
Q_conduction = Q * V
Where V is the volume of the wall. V = A * L.
So, V = 0.1 m² * 0.1 m = 0.01 m³
Now we have Q_conduction = (3 x 10^16 W/m³) * (0.01 m³) = 3 x 10^14 W
Substituting the values into the equation for ΔT:
ΔT = (Q_conduction * L) / (k * A)
ΔT = (3 x 10^14 W * 0.1 m) / (25 W/m°C * 0.1 m²)
ΔT = 1.2 x 10^13 °C
Now that we have ΔT, we can find the maximum temperature in the wall:
T_max = T_f + ΔT
T_max = 90 °C + 1.2 x 10^13 °C
T_max = 1.2 x 10^13 °C
Thus, the maximum temperature in the wall is 1.2 x 10^13 °C.